Chapter 6: X-rays

Example 6.1, Page 369

In [1]:
from math import *

#Variable declaration
i = 2e-003;  # Current through X-ray tube, A
e = 1.6e-019;  # Charge on an electron, C
V = 12.4e+003;  # Potential difference applied across X-ray tube, V 
m0 = 9.1e-031;  # Rest mass of the electron, Kg 

#Calculations&Results
n = i/e;    # Number of electrons striking the target per second
print "The number of electrons striking the target per sec = %4.2e electrons"%n
v = sqrt(2*e*V/m0);  # Velocity of the electrons, m/s
print "The speed with which electrons strike the target = %4.2e m/s"%v

The number of electrons striking the target per sec = 1.25e+16 electrons
The speed with which electrons strike the target = 6.60e+07 m/s

Example 6.2, Page 370

In [2]:
from math import *

#Variable declaration
e = 1.6e-019;  # Charge on an electron, C
V = 13.6e+003;  # Potential difference applied across X-ray tube, V 
m0 = 9.1e-031;  # Rest mass of the electron, Kg 

#Calculations
v = sqrt(2*e*V/m0);  # Velocity of the electron, m/s 

#Result
print "The maximum speed with which the electrons strike the target = %4.2e m/s"%v

The maximum speed with which the electrons strike the target = 6.92e+07 m/s

Example 6.3, Page 370

In [3]:
from math import *

#Variable declaration
d = 2.82e-010;  # Spacing of the rock-salt, m 
n = 2;  # Order of diffraction

#Calculations
theta = pi/2;    # Angle of diffraction, radian
# Braggs equation for X-rays of wavelength lambda is n*lambda = 2*d*sin(theta), solving for lambda
lamda = 2*d*sin(theta)/n;    # Wavelength of X-ray using Bragg's law, m

#Result
print "The longest wavelength that can be analysed by a rock-salt crystal = %4.2f angstrom"%(lamda/1e-010)

The longest wavelength that can be analysed by a rock-salt crystal = 2.82 angstrom

Example 6.4, Page 371

In [18]:
from math import *

#Variable declaration
lamda = 3e-011;  # Wavelength of the X-ray, m
d = 5e-011;  # Lattice spacing, m 

#Calculations&Results
# Bragg's equation for X-rays of wavelength lambda is n*lambda = 2*d*sin(theta), solving for thetas
for n in range(2,4):
    theta = degrees(asin((n*lamda)/(2*d)));    
    print "For n = %d, theta = %.1f degrees"%(n, theta)

For n = 2, theta = 36.9 degrees
For n = 3, theta = 64.2 degrees

Example 6.5, Page 371

In [19]:
from math import *

#Variable declaration
lamda = 3.6e-011;  # Wavelength of X-rays, m
n = 1;    # Order of diffraction
theta = 4.8;    # Angle of diffraction, degrees

#Calculations
# Braggs equation for X-rays is n*lambda = 2*d*sin(theta), solving for d
d = n*lamda/(2*sin(theta*pi/180));    # Interplanar spacing, m

#Result
print "The interplanar separation of atomic planes in the crystal = %4.2f angstrom"%(d/1e-010)

The interplanar separation of atomic planes in the crystal = 2.15 angstrom

Example 6.6, Page 371

In [20]:
#Variable declaration
lambda1 = 0.71;  # Wavelength of k alpha line in molybdenum, angstrom
Z1 = 42;        # Atomic number of Mo
Z2 = 29;        # Atomic number of Cu

#Calculations
# Wavelength of characteristic X-ray for K-alpha spectral line is given by 
# 1/lambda = 3/4*R*(Z-1)^2 then
lambda2 = lambda1*(Z1-1)**2/(Z2-1)**2;    # The wavelength of K alpha radiation in copper, m

#Result
print "The wavelength of K-alpha radiation in copper = %4.2f angstrom"%lambda2

The wavelength of K-alpha radiation in copper = 1.52 angstrom

Example 6.7, Page 372

In [56]:
from math import *

#Variable declaration
phi = pi/2;    # Scattering angle, degrees
m0 = 9.1e-031;  # Rest mass of an electron, kg
h =  6.62e-034;  # Planck's constant, J-s
c = 3e+008;  # Speed of light in vacuum, m/s 
E = 8.16e-014;  # Energy of gamma radiation, J

#Calculations
lamda = h*c/(E*1e-010);    # Wavelength of incident photon, angstrom 
lambda_prime = lamda+h*(1-cos(phi*pi/180))/(m0*c*1e-010);    # Wavelength of scattered photon, angstrom

#Result
print "The wavelength of radiation at 90 degrees = %6.4f angstrom"%(lambda_prime+lamda)

The wavelength of radiation at 90 degrees = 0.0487 angstrom

Example 6.8, Page 372

In [54]:
from math import *

#Variable declaration
phi = 90;    # Scattering angle, radian
m0 = 9.1e-031;  # Rest mass of the electron, kg
h =  6.62e-034;  # Planck's constant, J-s
c = 3e+008;  # Speed of light in vacuum, m/s 
lamda = 1.00 ;  # Wavelength of incident photon,in angstrom

#Calculations
del_lambda = (h*(1-round(cos(degrees(phi))))/(m0*c))/10**-10;    # Compton shift, angstrom

#Result
print "The Compton shift = %.4f angstrom"%del_lambda

The Compton shift = 0.0242 angstrom

Example 6.9, Page 373

In [55]:
from math import *

#Variable declaration
phi = pi/2;       # Scattering angle, radian
m0 = 9.1e-031;  # Rest mass of the electron, kg
h =  6.62e-034;  # Planck's constant, J-s
c = 3e+008;  # Speed of light in vacuum, m/s 

#Calculations
# As Compton shift = del_lambda = lambda, so
lamda = h*(1-cos(phi))/(m0*c*1e-010);    # Wavelength of incident photon, angstrom

#Result
print "The wavelength of incident radiation = %6.4f angstrom"%lamda

The wavelength of incident radiation = 0.0242 angstrom